racinnut15xm
Member
I had one and ordered another. Both are .420-421. Specs online show .425 are they all made by one company and all small?
they advertise it as 425 because that is the advertised size of the plate class that is allowed to race. That doesn't mean the plate is 425. It just means that if you want to race in that 425 class they'll sell you a 425 plate.they aren't selling you a 425 size plate they're selling you a plate for the 425 classI mean its advertised at .425 and it measures .420. Trying to eeek every last bit out of these things and I'm sure it will make a little difference. I didn't know if only one company was allowed to make them. I will contact dover on the next one.
I mean its advertised at .425 and it measures .420. Trying to eeek every last bit out of these things and I'm sure it will make a little difference. I didn't know if only one company was allowed to make them. I will contact dover on the next one.
Huh?You do realize you're talking .005" of an inch? and because it is a circle, its actually only .0025" of the diameter. If you plan to get a perfect plate, I'd be getting a certified No-Go tech tool to check with and not your harbor freight calipers...
His plate is .005" smaller than MAX. Since it is a round hole, the actual difference is .0025" overall circumference is it not? Maybe I'm not typing it out correctly, but I know what I'm trying to say! LOLHuh?
No you have that wrong, the diameter is .005" smaller than max. That's it, no other math is required, unless you are talking about the area of the hole. The circumference has nothing to do with the conversation of the hole being small.
We could talk about a % change in cross sectional area of the hole and that would be meaningful.
Where area = Pi x R^2
So the ideal area, right at limit (.425") would be 3.14159 x ((.425/2)x(.425/2)) = .14186 square inches.
The area of the plate you have (.420") would be 3.14159 x ((.420/2)x(.420/2)) = .13854 square inches.
The difference in area is .14186 - .13854 = .00318 square inches
The difference expressed as a percentage is .00318/.14186 x 100%= 2.3%
I like the % number that's how I like to think of things when we get on the dyno. .2 hp doesn't seem like much on the dyno but you start stacking a couple of those together and you've got something. These things produce such low numbers that a possible 2% increase in plate size would give a potential for mo powa. But alas Dover doesn't do this with the plates anymore. Back to the drawing boardNo you have that wrong, the diameter is .005" smaller than max. That's it, no other math is required, unless you are talking about the area of the hole. The circumference has nothing to do with the conversation of the hole being small.
We could talk about a % change in cross sectional area of the hole and that would be meaningful.
Where area = Pi x R^2
So the ideal area, right at limit (.425") would be 3.14159 x ((.425/2)x(.425/2)) = .14186 square inches.
The area of the plate you have (.420") would be 3.14159 x ((.420/2)x(.420/2)) = .13854 square inches.
The difference in area is .14186 - .13854 = .00318 square inches
The difference expressed as a percentage is .00318/.14186 x 100%= 2.3%